Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.
罗马数字有如下符号:
基本字符 | I | V | X | L | C | D | M |
对应阿拉伯数字 | 1 | 5 | 10 | 50 | 100 | 500 | 1000 |
计数规则:
- 相同的数字连写,所表示的数等于这些数字相加得到的数,例如:III = 3
- 小的数字在大的数字右边,所表示的数等于这些数字相加得到的数,例如:VIII = 8
- 小的数字,限于(I、X和C)在大的数字左边,所表示的数等于大数减去小数所得的数,例如:IV = 4
- 正常使用时,连续的数字重复不得超过三次
- 在一个数的上面画横线,表示这个数扩大1000倍(本题只考虑3999以内的数,所以用不到这条规
思路 1 : 从后往前遍历, 遇到一样的就粘贴上。 i = i + 2; num = num / 10;
思路2 : 只有4,9是左减,其余的数字都是右加。所以,构建两个数组,分别是数字和对应的字母。数字,要把所有的4,9加进来。包括4,9,40,90,400,900.
从前往后, 用取余的方法 来进位 ,num %= base[i];
public class Solution {
public String intToRoman(int num) {
if(num <= 0 || num >= 4000)
return null;
String roman[] = { "I","V","X","L","C","D","M" };
String RomanString = "";
int i = 0;
while(num != 0)
{
int temp = num % 10;
String Rstr = "";
switch(temp)
{
case 1:
Rstr = roman[i];
break;
case 2:
Rstr = roman[i] + roman[i];
break;
case 3:
Rstr = roman[i] + roman[i] + roman[i];
break;
case 4:
Rstr = roman[i] + roman[i+1];
break;
case 5:
Rstr = roman[i+1];
break;
case 6:
Rstr = roman[i+1] + roman[i];
break;
case 7:
Rstr = roman[i+1] + roman[i] + roman[i];
break;
case 8:
Rstr = roman[i+1] + roman[i] + roman[i] + roman[i];
break;
case 9:
Rstr = roman[i] + roman[i+2];
break;
}
RomanString = Rstr + RomanString;
i = i + 2;
num = num / 10;
}
return RomanString;
}
}
public class Solution {
public String intToRoman(int num) {
int[] base = { 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
String[] roman = { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X",
"IX", "V", "IV", "I" };
String result = "";
for (int i = 0; num > 0; i++) {
int count = num / base[i];
num %= base[i];
while (count > 0) {
result += roman[i];
--count;
}
}
return result;
}
}